Question 196070
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Given Quadratic Eqn, {{{x^2-5x-14=0}}}


X-Intercept of the vertex,
{{{x=-b/2a=-(-5)/(2*1)=red(5/2)}}}


Y-Intercept, substitute {{{x=5/2}}} or {{{x=2.5}}}
{{{y=(5/2)^2-5(5/2)-14}}}
{{{y=6.25-12.5-14}}}
{{{red(y=-20.25)}}}



For the x-intercepts:

Discriminant: {{{b^2-4ac=-5^2-4(1)(-14)=25+56=81}}}
Then,
{{{x=(-(-5)+-sqrt(81))/(2*1)=(5+-9)/2}}}
{{{x=(5+9)/2=14/2=red(7)}}}
{{{x=(5-9)/2=-4/2=red(-2)}}}

{{{drawing(400,400,-8,12,-22,10,grid(1),graph(400,400,-8,12,-22,10,x^2-5x-14),blue(circle(5/2,-20.25,.15)),green(circle(7,0,.15)),green(circle(-2,0,.15)))}}}


Thank you,
Jojo</font>